What is the weight of the cargo removed from the steamer if its draft has decreased

What is the weight of the cargo removed from the steamer if its draft has decreased by 20 cm? The horizontal section of the steamer at water level is 4000 m.

Δh = 20 cm = 0.2 m.

g = 9.8 m / s2.

S = 4000 m2.

ρ = 1000 kg / m3.

Δm -?

Two forces act on the steamer in the water: the force of gravity Ft directed vertically downward, and the buoyancy force of Archimedes Farch directed vertically upward.

Ft = Farch.

The force of gravity is determined by the formula: Ft = m * g, where m is the mass of the steamer, g is the acceleration of gravity.

The buoyancy force of Archimedes Farch is determined by the formula: Farch = ρ * g * V. Where ρ is the density of the liquid in which the body is immersed, g is the acceleration of gravity, V is the volume of the immersed part of the body in the liquid.

m1 * g = ρ * g * V1.

m1 = ρ * V1.

m2 * g = ρ * g * V2.

m2 = ρ * V2.

Δm = m1 – m2 = ρ * V1 – ρ * V2 = ρ * (V1 – V2) = ρ * S * Δh.

Δm = 1000 kg / m3 * 4000 m2 * 0.2 m = 800000 kg.

Answer: the cargo was removed from the ship Δm = 800,000 kg.